14-5w=w^2

Solution for 14-5w=w^2 equation:

14-5w=w^2

We move all terms to the left:

14-5w-(w^2)=0

determiningTheFunctionDomain
-w^2-5w+14=0

We add all the numbers together, and all the variables

-1w^2-5w+14=0

a = -1; b = -5; c = +14;
Δ = b2-4ac
Δ = -52-4·(-1)·14
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-9}{2*-1}=\frac{-4}{-2}
=+2
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+9}{2*-1}=\frac{14}{-2}
=-7
$